Displaystyle sin theta sqrt-1-tan -2-theta -displaystyle sin theta cos theta displaystyle tan theta displaystyle cot theta 1

The correct option is B tan-x3B8-Use 1 -xA0-tan2-x3B8-sec2-x3B8-

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Question

$sin θ \sqrt{1 + {tan}^{2} θ} =$
$sin θ cos θ$
$tan θ$
$cot θ$
1

tan θ

sin θ cos θ

cot θ

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Prave that:

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More than One Answer Type
एक से अधिक उत्तर प्रकार के प्रश्न

If

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

and

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

where 0° < θ < 90°, then the correct option(s) is/are

यदि

x = (\frac{{sin}^{2} θ {sec}^{2} θ + 1}{{sec}^{2} θ}) + (\frac{{cot}^{2} θ}{1 + c o s e c θ}),

y = \frac{1 + {cos}^{2} θ c o s e c^{2} θ}{c o s e c^{2} θ} + \frac{{tan}^{2} θ}{1 + sec θ}

तथा

z = \frac{1 - {sin}^{2} θ + sin θ cos θ}{sin θ - {sin}^{3} θ},

जहाँ 0° < θ < 90°, तब सही विकल्प है/हैं

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